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Author Topic: What happens if I dangle a string over a black hole's event horizon?  (Read 10036 times)

Offline mlandri

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Hi all,

I'm trying to wrap my brain around black hole event horizons and would appreciate some help.

Let's say that I head off to the nearest black hole with my good friend Bob and a long, very strong string. For the sake of simplicity, assume that the black hole is not spinning. I hold one end of the string on the safe side of the event horizon and send off Bob to cross the event horizon with the other end of the string. Bob repeatedly and regularly tugs on his end of the string as he moves toward and eventually crosses over the black hole event horizon? Every time I feel a tug, I give Bob a thumbs-up.

Ok. So what happens?

From my perspective, Bob's tugging will slow and eventually stop altogether. Right? From Bob's perspective, on the other hand, he happily crosses over the event horizon tugging away without noting anything unusual. Both perspectives are supposed to be equally valid. What I can't figure out is what Bob will see as he watches for my and the string's reaction to his repeated tugs as he approaches and eventually crosses the event horizon.

One thought that I had was that if I intended to hold my end of the string forever, then this in conjunction with the black hole's gravitational pull would force me to choose one of the following or (similar) courses of action.

1) Allow the black hole to pull the string and me along with it past its event horizon

2) Pull back on the string and resist the black hole's gravity until the string brakes

3) Drop the string and head back home for a cold beer, allowing the string to follow Bob past the event horizon

From Bob's perspective, due to the effects of time dilation, I would apparently make this decision (at the latest) as he found himself crossing the black hole's event horizon. Hence, Bob wouldn't have a chance to tug on the string again while we were on separate sides of the event horizon and still attached by the string.

Does this sound reasonable, or am I completely off base? If this the correct, then it would seem that Bob would have to perceive time in the rest of the visible universe as exponentially accelerating as he approached the black hole's singularity (and before he was pulled apart into little Bob bits).

Thanks!


[MOD EDIT - PLEASE PHRASE YOUR THREAD TITLES AS QUESTIONS. THANKS. CHRIS]
« Last Edit: 25/10/2010 18:32:23 by chris »


 

Offline JP

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Hence, Bob wouldn't have a chance to tug on the string again while we were on separate sides of the event horizon and still attached by the string.

I think that's at least part of the problem.  You can't be on opposite sides of the even horizon and still be attached.  The problem is that nothing can cross the event horizon, and this includes the forces holding the string together.  The string would basically have to break apart at the event horizon, since the forces can't hold it together at that point.
 

Offline Soul Surfer

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What happens depends a great deal on the size of the black hole for a stellar mass black hole the gravitational gradients are far too great and everything would be ripped apart long before it got anywhere near the horizon.  If it was a quiet multi billon solar mass black hole two attached bodies orbiting the hole (you would have to be orbiting the hole to provide a platform  from which to pull) could in theory be on either side of the horizon a  the same time and if you pulled the string whether you dropped in or he came out would be dependant on the total energy of thew system.  however neither of you would see or experience anything significantly different because to both of you  would be just like so much empty space with a distorted view of the rest of the universe  the only difference is that bob would not be able to see the stars (but he would be able to see you quite normally)  and you would have a very restricted view of them
 

Offline mlandri

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JP, SS, Thanks for the replies...

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You can't be on opposite sides of the even horizon and still be attached.

Assuming that the string and Bob could survive their trip to the event horizon, I was under the impression that this could, in fact, happen. Conditions on one side of the event horizon should not be that much different from those on the other. The main difference is that space-time is falling toward the singularity at or faster than the speed of light at and behind the black hole's event horizon. Space-time is traveling near but less than the speed of light toward the black hole's singularity close to but on the other side of its event horizon. The gravitational gradient and its growing effects on space-time would not be apparent over short distances even while crossing over the event horizon (well, until you get much closer to the singularity). This is the conundrum. How/Why can one cross such a threshold and not notice?

It seems to me that time dilation must be the primary reason why Bob could cross the black hole's event horizon but not notice that he could no longer interact with the rest of the Universe (or me). Of course, I could have this completely wrong. I am no physicist :)

Quote
if you pulled the string whether you dropped in or he came out would be dependant on the total energy of thew system.

SS: Would pulling Bob out be possible? Wouldn't that be the equivalent of accelerating him faster than the speed of light?

« Last Edit: 05/09/2010 03:39:21 by mlandri »
 

Offline Soul Surfer

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mlandri  this is a totally theoretical problem because no materials could ever be strong enough or light enough to hold together particularly when you consider the dimensions involved.  The event horizon of a one billion solar mass black hole if placed in the solar system extends out to the orbit of Uranus  about 20 times the distance of the earth from the sun.  The last stable orbit around it is as much again further away about 40 times the distance of the earth (near the orbit of neptune) from the sun further out so the "string" would have to be  two thousand million miles long !

It is vitally important tho get the scales of things properly in perspective.  remember too that even a black hole this big with a diameter of a few light hours is absolutely tiny compared with the typical distance between stars of light years.

You are right when you say you could cross the event horizon of a big black hole without noticing it because the gravity gradient at the EH is tiny for a one billion solar mass hole.

The smallest black hole a human being could in theory get into without suffering excessive stress  say no more than one earth gravity between head and feet,  is about the size of the planet Jupiter and has 50,000 solar masses in it.

http://xaonon.dyndns.org/hawking/   gives you a very good picture of the relative sizes and fields of black holes
 

Offline Bored chemist

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Crossing the event horizon is a bit like climbing into space from earth.
If I want to leave the earth ballisticaly, I need to go faster than the escape velocity. On the other hand, if I just make a long ladder, I can leave at any speed I like.
For a black hole the escape velocity is greater than C but if I have a rope to climb, I don't need to exceed that velocity.
 

Offline Soul Surfer

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Rereading your initial question I will put my first answer in a different way which may help. 

firstly forget the string and chat to Bob via a radio link and accept that Bob (poor sucker) is doomed.  This radio link will be very adaptable and cope with the (very large) doppler shifts caused by the gravitational fields and may need some voice processing to keep the speech intelligible. 

You and Bob will take up a position orbiting just outside the last stable orbit at a speed just below the speed of light  you also have a mate fred who is a long way away in a very safe place and also in radio contact with you as you orbit the hole.  Bob will then launch off and using a lot of rocket power will keep stationed in line between you and the centre of the hole.  Let us also discount the communications time because we are doing this with a Jupiter size black hole and time delays will only be a few seconds (even though the accelerations involved may cause problems)

As Bob approaches the event horizon of the hole as seen by Fred. Fred tells you he has seen bob cross the horizon and vanish.
 
Forget about this taking an infinite time because its just like saying that the temperature of a glass of hot tea never cools down to room temperature it is a decay function and just an idea of the Gee Whizz  Science Crew.

You however are still chatting happily to Bob although he reports that the stars in your direction have now all vanished but he can still see you and he is feeling a bit stretched out like he is hanging from a bar.  This is because the event horizon for you is in fact closer to the hole because you are deep in the gravitational field of the hole.  Bob looks towards the hole and in the last few seconds before communication breaks of and he crosses the event horizon for you, he reports that he can see nothing more in the direction of the hole except cold and dark.  This is  because the event horizon for him is a bit further ahead and will continue to be so until he gets very much closer to the real action but by then he will be will and truly spaghettified. 

You report this observation back to fred  who says thank you for taking part in the experiment and that the theory was properly confirmed and by the way there was not enough fuel for you to get back out from the your place near the last stable orbit.  have a good death over and out!

The true fact is that the event horizon is not a wall which once you get through all will be revealed and you can see what is inside the hole but a bit like the end  a rainbow, something that you can never reach and see beyond.  It is just the point where the gravitational field causes the escape velocity to infinity to become that of light. so as Bob is falling inside the event horizon to infinity he experience two visibility horizons one for what he can see inwards and one for what he can see looking outwards.  In general these are both expected to be black so it's not likely that he will see much if the hole is quiescent.






« Last Edit: 05/09/2010 22:45:03 by Soul Surfer »
 

Offline yor_on

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---Quote---

A misconception concerning event horizons, especially black hole event horizons, is that they represent an immutable surface that destroys objects that approach them. In practice, all event horizons appear to be some distance away from any observer, and objects sent towards an event horizon never appear to cross it from the sending observer's point of view (as the horizon-crossing event's light cone never intersects the observer's world line). Attempting to make an object approaching the horizon remain stationary with respect to an observer requires applying a force whose magnitude becomes unbounded (becoming infinite) the closer it gets.

For the case of an horizon perceived by a uniformly accelerating observer in empty space, the horizon seems to remain a fixed distance from the observer no matter how its surroundings move. Varying the observer's acceleration may cause the horizon to appear to move over time, or may prevent an event horizon from existing, depending on the acceleration function chosen. The observer never touches the horizon, and never passes a location where it appeared to be.

For the case of an horizon perceived by an occupant of a De Sitter Universe, the horizon always appears to be a fixed distance away for a non-accelerating observer. It is never contacted, even by an accelerating observer.

For the case of the horizon around a black hole, observers stationary with respect to a distant object will all agree on where the horizon is. While this seems to allow an observer lowered towards the hole on a rope to contact the horizon, in practice this cannot be done. If the observer is lowered very slowly, then, in the observer's frame of reference, the horizon appears to be very far away, and ever more rope needs to be paid out to reach the horizon. If the observer is quickly lowered by another observer, then indeed the first observer, and some of the rope can touch and even cross the (second observer's) event horizon. If the rope is pulled taut to fish the first observer back out, then the forces along the rope increase without bound as they approach the event horizon, and at some point the rope must break. Furthermore, the break must occur not at the event horizon, but at a point where the second observer can observe it.

Attempting to stick a rigid rod through the hole's horizon cannot be done: if the rod is lowered extremely slowly, then it is always too short to touch the event horizon, as the coordinate frames near the tip of the rod are extremely compressed. From the point of view of an observer at the end of the rod, the event horizon remains hopelessly out of reach. If the rod is lowered quickly, then the same problems as with the rope are encountered: the rod must break and the broken-off pieces inevitably fall in.

These peculiarities only occur because of the supposition that the observers be stationary with respect to some other distant observer. Observers who fall into the hole are moving with respect to the distant observer, and so perceive the horizon as being in a different location, seeming to recede in front of them so that they never contact it. Increasing tidal forces (and eventual impact with the hole's gravitational singularity) are the only locally noticeable effects. While this seems to allow an in-falling observer to relay information from objects outside their perceived horizon but inside the distant observer's perceived horizon, in practice the horizon recedes by an amount small enough that by the time the in-falling observer receives any signal from farther into the hole, they've already crossed what the distant observer perceived to be the horizon, and this reception event (and any retransmission) can't be seen by the distant observer.

---End of quote----

From Event horizon 

" Hawking predicted that energy fluctuations from the vacuum causes the generation of particle-antiparticle pairs near the event horizon of the black hole. One of the particles falls into the black hole while the other escapes, before they have an opportunity to annihilate each other. The net result is that, to someone viewing the black hole, it would appear that a particle had been emitted.

Since the particle that is emitted has positive energy, the particle that gets absorbed by the black hole has a negative energy relative to the outside universe. This results in the black hole losing energy, and thus mass (because E = mc2).

Smaller primordial black holes can actually emit more energy than they absorb, which results in them losing net mass. Larger black holes, such as those that are one solar mass, absorb more cosmic radiation than they emit through Hawking radiation.  "


And conservation_of_energy

(The last one I just let follow because I have read too many saying that black holes 'communicate'. I don't agree to that view myself as it would mean that it no longer would be a singularity. And i like the explanation, it makes perfect sense to me, and as far I can see does not involve any communication, if you don't see the particle-pairs as entangled and then also assume that it is the one destroyed that somehow will imprint the information on its entangled surviving twin.) And that's why I find conservation of energy interesting too.
« Last Edit: 08/09/2010 09:35:53 by yor_on »
 

Offline mlandri

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yor_on. Thank you for finding that answer! I think that it gets to the heart of my question.

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For the case of the horizon around a black hole, observers stationary with respect to a distant object will all agree on where the horizon is. While this seems to allow an observer lowered towards the hole on a rope to contact the horizon, in practice this cannot be done. If the observer is lowered very slowly, then, in the observer's frame of reference, the horizon appears to be very far away, and ever more rope needs to be paid out to reach the horizon. If the observer is quickly lowered by another observer, then indeed the first observer, and some of the rope can touch and even cross the (second observer's) event horizon. If the rope is pulled taut to fish the first observer back out, then the forces along the rope increase without bound as they approach the event horizon, and at some point the rope must break. Furthermore, the break must occur not at the event horizon, but at a point where the second observer can observe it.

SS. Thanks for your follow-up. Had to ponder the responses a bit.

I think that I am tracking now. Not sure that I sufficiently appreciated Einstein before now :)
 

Offline Bill S

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Quote from: SS
as Bob is falling inside the event horizon to infinity he experience two visibility horizons one for what he can see inwards and one for what he can see looking outwards.

Thanks, folks, for a good discussion which has already contributed much to my limited understanding of BHs. Given time to digest it, I'm sure there will be lots of questions, but here is one to start.  Why is Bob unable to see out? I didn't realise that light had any trouble getting in through the EH.
 

Offline Soul Surfer

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The inside of a black hole is like a mirror and "reflects" any light falling on it from the inside.  Light falling on the outside of the hole is severely distorted and scrambled.  Looking away from the centre there is one undistorted ray that is a straight line producing a small image that is relatively undistorted but this is only true if the black hole is not rotating. If it is rotating(almost certain), frame dragging completely scrambles any light coming in from outside rays from other directions are severely distorted and can come at you from all sorts of directions.  If it is rotating, frame dragging completely scrambles any light coming in from outside  except possibly if you happen to be placed exactly at one of the polar axes.  remember also there is a point close to the event horizon where light can go into orbit go round a few times and then either fall in or come out of the hole.
 

Offline Bill S

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Quote from: mlandri
The main difference is that space-time is falling toward the singularity at or faster than the speed of light at and behind the black hole's event horizon.

Can space-time be said to be "falling", rather than just curved? I would have assumed that this was just a figure of speech had it not been accorded a speed.
 

Offline Bill S

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Did I kill this thread?  If I did, I offer a big grovelling apology.  Most of the grovelling being due to the fact that I have lots more questions about black holes, and where better to look for answers?  Here are a couple of those questions.

Quote from: SS
The event horizon of a one billion solar mass black hole if placed in the solar system extends out to the orbit of Uranus  about 20 times the distance of the earth from the sun.

In the absence of any observer with reference to whom the E H might appear to move, would there be a specific point (e.g. a spherical surface)within that vast distance beyond which light would not be able to escape?

Quote
The smallest black hole a human being could in theory get into without suffering excessive stress  say no more than one earth gravity between head and feet,  is about the size of the planet Jupiter and has 50,000 solar masses in it.

"Into", but presumably not to the centre?

 

Offline jartza

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Crossing the event horizon is a bit like climbing into space from earth.
If I want to leave the earth ballisticaly, I need to go faster than the escape velocity. On the other hand, if I just make a long ladder, I can leave at any speed I like.
For a black hole the escape velocity is greater than C but if I have a rope to climb, I don't need to exceed that velocity.

The climbing exercise of climbing out of an event horizon is such that you lose all your weight.

climbing energy = mc˛

« Last Edit: 25/10/2010 09:20:06 by jartza »
 

Offline Bill S

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Quote from: jartza
The climbing exercise of climbing out of an event horizon is such that you lose all your weight.

Why?
 

Offline jartza

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Quote from: jartza
The climbing exercise of climbing out of an event horizon is such that you lose all your weight.

Why?

A person weighting 80 kilograms must use 80 kilograms of energy when climbing up the ladders that lead up from the event horizon. And a person weighting 80 kilograms has about 0 kilograms of energy and 80 kilograms of matter, which converted into energy makes 80 kilograms of energy.



 

Offline jartza

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When a person is climbing up from event horizon, gravitometers at upper position do not detect any change happening.

When a person is lowered into a event horizon, first gravitometers detect that some mass-energy is moving downwards, then later gravitometers detect that mass-energy is moving upwards.

You see, when 1 kg of mass-energy is lowered, the mass-energy of the lowering device increases by 1 kg, so in the end same amount of mass-energy is at same location as  in the beginning.
 

Offline Bill S

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Thanks Jartza.  Threads do tend to meander (in an interesting way, of course) and looking back through this one I find several unanswered questions.  I'm sure someone out there has the answers.  Perhaps there is a fear that answering them will encourage me to ask lots more!  Such a fear would be well founded.  :P
 

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